Problem: Let $R$ be the region enclosed by the $x$ -axis, the $y$ -axis, the line $y=3$, and the curve $y=\text{ln}(x)$. $y$ $x$ ${y=\text{ln}(x)}$ $ R$ $ 3$ $(0,0)$ A solid is generated by rotating $R$ about the $y$ -axis. Which one of the definite integrals gives the volume of the solid? Choose 1 answer: Choose 1 answer: (Choice A) A $\pi\int_0^{e^3} \left[\text{ln}(y)\right]^2dy$ (Choice B) B $\pi\int_0^3 \left[\text{ln}(y)\right]^2dy$ (Choice C) C $\pi\int_0^{\text{ln}(3)} e^{2y}dy$ (Choice D) D $\pi\int_0^3 e^{2y}dy$
Solution: Let's imagine the solid is made out of many thin slices. $y$ $x$ ${y=\text{ln}(x)}$ Notice the slices are horizontal, because we are rotating $R$ about the $y$ -axis. Each slice is a cylinder. Let the thickness of each slice be $dy$ and let the radius of the base, as a function of $y$, be $r(y)$. Then, the volume of each slice is $\pi [r(y)]^2\,dy$, and we can sum the volumes of infinitely many such slices with an infinitely small thickness using a definite integral: $\int_a^b \pi [r(y)]^2\,dy$ This is called the disc method. What we now need is to figure out the expression of $r(y)$ and the interval of integration. Let's consider one such slice. $y$ $x$ ${y=\text{ln}(x)}$ $ 3$ $(0,0)$ $r$ The radius is equal to the distance between the curve $y=\text{ln}(x)$ and the $y$ -axis. To find it, we need to solve the equation for $x$ : $x=e^y$ So, for any $y$ -value, $r(y)=e^y}$. Now we can find an expression for the area of the cylinder's base: $\begin{aligned} &\phantom{=}\pi [r(y)}]^2 \\\\ &=\pi\left[e^y}\right]^2 \\\\ &=\pi e^{2y} \end{aligned}$ The bottom endpoint of $R$ is at $y=0$ and the top endpoint is at $y=3$. So the interval of integration is $[0,3]$. Now we can express the definite integral in its entirety! $\begin{aligned} &\phantom{=}\int_0^3 \pi e^{2y}dy \\\\ &=\pi\int_0^3 e^{2y}dy \end{aligned}$